Let\ f(x)=x^2-4x-2, x≠2 3, x=2 Which statement about the continuity of f at x=2 is correct?

f has a removable discontinuity at x=2.

f has a jump discontinuity at x=2.

f is discontinuous at every point in [0,4].

Let\ $f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 4}{x - 2}, 3, x \neq = 2 x = 2$ Which statement about the continuity of $f$ at $x = 2$ is correct?

easy MCQ 5 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

f

is continuous at

x = 2

The correct answer is $f$ is continuous at $x = 2$ . — Therefore $f$ is NOT continuous at $x = 2$ .

Compute the limit of $f (x)$ as $x \to 2$ . Since $x \neq = 2$ , $\frac{x ^{2} - 4}{x - 2} = \frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2$ .
Take the limit: $lim_{x \to 2} (x + 2) = 4$ .
Given that $f (2) = 3$ , the limit $4$ does not equal $f (2) = 3$ .
Therefore $f$ is NOT continuous at $x = 2$ .

Common mistake: Assuming algebraic simplification gives correct value at the point of discontinuity.

Concept tested: Continuity of piecewise function